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The value of $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$ is -

JEE · Math · previous-year question

  1. A.$${1 \over {256}}$$
  2. B.$${1 \over {2}}$$
  3. C.$${1 \over {1024}}$$
  4. D.$${1 \over {512}}$$correct

Answer

D. $${1 \over {512}}$$

Explanation

Given $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$ Let $${\pi \over {{2^{10}}}}\, = \,\theta $$ $$ \therefore $$ $${\pi \over {{2^9}}}\, = \,2\theta $$ $${\pi \over {{2^8}}}\, = \,{2^2}\theta $$ $${\pi \over {{2^7}}}\, = \,{2^3}\theta $$ . . $${\pi \over {{2^2}}}\, = \,{2^8}\theta $$ So given term becomes, $$\cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta $$$$.\sin {\pi \over {{2^{10}}}}$$ = $$(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}$$ = $${{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}$$ = $${{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}$$ = $${{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}$$ = $${1 \over {{2^9}}}$$ = $${1 \over {512}}$$ Note : $$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$$ = $${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$$

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